Timeline for Whitney sum formula for topological Pontryagin classes
Current License: CC BY-SA 4.0
12 events
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| May 23, 2020 at 12:52 | comment | added | archipelago | $[X,BSO_\mathbb{Q}]\rightarrow [X,BSTOP_\mathbb{Q}]$ is an isomorphism. The rational Pontryagin classes of $x\in [X,BSTOP]$ only depend on its image in $[X,BSTOP_\mathbb{Q}]$. | |
| May 23, 2020 at 12:25 | comment | added | Igor Belegradek | If $K_Q$ is the product of $K(4i,\mathbb Q)$ spaces over all integers $i>0$, then all we have is an isomorphism $[BTOP, K_Q]\to [BO, K_Q]$ which sends the topological Pontryagin class to the usual one. | |
| May 23, 2020 at 12:22 | comment | added | Igor Belegradek | @archipelago and@Oscal: The edit does not address my confusion. I do not understand the claim "rationally, every topological bundle has a unique vector bundle structure". If say $X$ is a finite complex, the cokernel $[X,BO]\to [X,BTOP]$ is finite. The inclusion could be like $2\mathbb Z\subset\mathbb Z$. If $x\in [X,BTOP]$, which vector bundle is $x$ equivalent to rationally? If $t\in [X, BTOP]$ has order $n$, i.e. $nt=0$, how to we know that $p(t)=1$? Why is $p(x+t)=p(x)p(t)$? Here $p=1+p_1+p_2+\dots$, the total Pontryagin class. | |
| May 23, 2020 at 9:35 | history | edited | Oscar Randal-Williams | CC BY-SA 4.0 |
added 425 characters in body
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| May 23, 2020 at 9:20 | comment | added | archipelago | @Igor Belegradek: stably and rationally, every topological bundle has a unique vector bundle structure; this is just rephrasing that the map BSO->BSTOP is a rational equivalence. The fact that this map is one of H-spaces says exactly that the induced vector bundle structure is compatible with sums. Note that it suffices to know that BSO->BSTOP is a rational equivalence; you do not need to use that the map to the product of the K(Z,4i)s is one (however, to prove the former, one needs the latter). | |
| May 22, 2020 at 22:49 | comment | added | Igor Belegradek | @Oscar: actually, I am confused. It seems your argument only proves that topological Pontryagin classes satisfy the Whitney sum formula on vector bundles. I need it for topological bundles. Basically, there are two H-maps which are rational equivalences from $BSO$: one to $BSTOP$ and the other one to the product of $K(\mathbb Z, 4i)$ spaces where $i$ varies over positive integers. The latter map is the integral Pontryagin class. One needs to show that the topological Pontryagin class completes the square rationally. None of the maps from $BSO$ can be reversed, I think. | |
| May 22, 2020 at 14:20 | vote | accept | Igor Belegradek | ||
| May 22, 2020 at 13:35 | comment | added | Connor Malin | Sorry, I deleted my comment in fear that I was missing something. For posterity, I said that usually a group will not be an infinite loop space. | |
| May 22, 2020 at 13:05 | comment | added | Igor Belegradek | @ConnorMalin: thank you, I see now that Boardman-Vogt's result has quite a few technical assumptions; they speak of symmetrical monoidal functors. I gather, the point is that Whitney sum is commutative. | |
| May 22, 2020 at 12:29 | comment | added | Igor Belegradek | Is it true more generally that for any topological monoid (or at least group) homomorphism $G\to H$ the corresponding map of classifying spaces $BG\to BH$ is an infinite loop map? | |
| May 22, 2020 at 12:29 | comment | added | Igor Belegradek | Thank you! For my record here is a reference: on p.215 of Boardman-Vogt's book "Homotopy invariant algebraic structures on topological spaces" the authors say (roughtly) that since $O\to Top\to F$ are topological monoid homomorphisms under the Whitney sum, these maps are infinite loop space maps, and hence so are the corresponding maps of classifying spaces $BO\to BTop\to BF$. Actually, instead of this last sequence they write $BO\to Top\to F$ which I assume is a misprint. | |
| May 22, 2020 at 7:05 | history | answered | Oscar Randal-Williams | CC BY-SA 4.0 |